# Pontriagin reflexivity of the character group

For an Abelian topological group $$G$$ by $$G^{\wedge}$$ we denote the Pontryagin dual of $$G$$, i.e. the group of continuous homomorphisms $$G\to\mathbb T:=\{z\in\mathbb C:|z|=1\}$$. The group $$G^{\wedge}$$ is endowed with the topology of uniform convergence on compact subsets of $$G$$. A topological group $$G$$ is called Pontryagin reflexive if the canonical homomorphism $$G\to (G^\wedge)^\wedge$$ is a topological isomorphism.

Problem. Let $$G$$ be a metrizable Abelian topological group. Is the Pontriagin dual $$G^{\wedge}$$ of $$G$$ Pontryagin reflexive?

(This problem was posed 21.09.2017 by Lydia Aussenhofer on page 71 of Volume 1 of the Lviv Scottish Book).

• Is it clear that $G^{\wedge}$ is metrizable? – YCor Sep 25 '17 at 22:40
• @YCor, maybe I miss the point of your question, but $\hat G$ doesn't have to be metrisable to ask whether it is reflexive, right? – LSpice Sep 25 '17 at 22:53
• If G is metrizable then $G^\wedge$ is a hemicompact $k$--space. This implies that the canonical mapping $G^\wedge\to G^{\wedge\wedge\wedge}$ is continuous (even an embedding). Only the surjectivity of this mapping is a problem. – LDA Sep 26 '17 at 9:06
• I hope you know that in the case when $G$ is a (metrizable) locally convex space the answer is "yes", see e.g. link.springer.com/article/10.1023%2FA%3A1020929201133 or ams.org/mathscinet-getitem?mr=2037513 – Sergei Akbarov Sep 26 '17 at 10:08
• In my previous comment I forgot to indicate that $G$ must be complete, I am sorry. – Sergei Akbarov Sep 26 '17 at 18:51

Not an answer to this problem.
For a non-reflexive $G$ see example: Exercise (23.32) in Hewitt & Ross, Abstract Harmonic Analysis I (Springer 1963).

Consider a topological vector space $L^p(\mathbb R)$, $0 < p < 1$. They show that the (separable metrizable) topological group $G = L^p(\mathbb R)$ with operation of addition has no nonzero continuous characters. So $\widehat{G} = \{0\}$ and $\widehat{\widehat{G}} = \{0\}$, not isomorphic to $G$.

Facts used (Hewitt & Ross include outlines of proofs):

(Day, 1940) this topological vector space has no nonzero continuous linear functionals

(Hewitt & Zuckerman, 1950) every continuous character $\chi$ on $G$ has the form $\chi(x) = \exp(2\pi i f(x))$ for some continuous linear functional $f$.

• This says that this $G$ isn't reflexive, but it doesn't seem to provide an example of a non-reflexive Pontryagin dual. I apologize if I'm missing something... – Alon Amit Sep 26 '17 at 0:17
• You are right. Not an answer to the original problem. – Gerald Edgar Sep 26 '17 at 0:19